Borda’s method is to let each voter rank the candidates by importance, then assign weights given by the rank position, to add the weights per candidate for all voters, and then select the candidate with the highest value. Note that the method appears sensitive to preference reversal, see below.

Condorcet’s method is to vote on all pairs of candidates, and to select the one who wins from all alternatives. Note that such a “Condorcet winner” does not need to exist. In that case the margins of winning can be used to solve the deadlock - but this increases the sensitivity to who participates.

The following example is taken from Saari (2001ab). Consider a budget of three candidates A, B and C, and let there be 114 voters. When we neglect indifference and use strict preference only, then with 3 candidates there are 3! = 6 possible ways of ranking them. Table 13 contains an arbitrary allocation of those voters over such preferences. The highest ranking candidate gets rankorder weight 3, the second gets weight 2, and the least preferred candidate gets weight 1. In the table we can read for example that there are 33 candidates with preference A > B > C.

Table 13: Voting example

Number of voters

Candidates and their rank order weight

Sum 114

A

B

C

33

3

2

1

0

3

1

2

25

2

1

3

17

1

2

3

14

1

3

2

25

2

3

1

Results of the procedures

Mostly preferred

33+0 = 33

14+25 = 39

25+17 = 42

Borda

230

242

212

Pairs: A vs B

58

56

-

A vs C

58

-

56

B vs C

-

72

42

The different voting schemes result into different decisions:

1) Plurality: Voters give one single vote to the candidate of their highest preference. For candidate A we consider its column, select the rows with the score 3, and add the associated numbers of voters 33 + 0 = 33. And so on. Candidate C gets most votes, namely 42.

2) Borda: The votes are weighted with the rank order weight. De column for A is multiplied row by row with the number of voters 3 * 33 + 3 * 0 + 2 * 25 + … = 230. Candidate B gets most votes, namely 242. (Scores -1, 0, 1 might calculate easier.)

3) Condorcet: Voting pairwise over A versus B, there are 33 + 0 + 25 = 58 voters who give A a higher rankorder than B. Etcetera. Candidate A appears to win from both B and C, and then is the “Condorcet winner”.

This example shows that A, B and C can all be winners, depending upon the method selected. The properties of the methods then are the true issue.